\(a,\widehat{aOn}+\widehat{mOn}+\widehat{mOb}=180^o\left(kề.bù\right)\\ \Leftrightarrow70^o+\widehat{mOn}+40^o=180^o\\ Vậy:\widehat{mOn}=180^o-\left(70^o+40^o\right)=70^o\\ b,Vì:\widehat{aOn}=\widehat{mOn}\\ Mà.tia.On.nằm.giữa.2.tia.Oa.và.Om.nên:\\ On.là.tia.phân.giác.góc.\widehat{aOm}\)

\(\dfrac{1}{4}-\left(2x+\dfrac{1}{2}\right)^2=0\)

        \(\left(2x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\) 

=>      \(\left(2x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{1}{2}\right)^2\) 

=>        \(2x+\dfrac{1}{2}=\pm\dfrac{1}{2}\) 

TH1: 

\(2x+\dfrac{1}{2}=\dfrac{1}{2}\)

        \(2x=\dfrac{1}{2}-\dfrac{1}{2}=0\) 

          \(x=0\) 

TH2:

\(2x+\dfrac{1}{2}=-\dfrac{1}{2}\)

       \(2x=-\dfrac{1}{2}-\dfrac{1}{2}\)  

        \(2x=-1\) 

            \(x=\dfrac{-1}{2}\) 

      Vậy \(x\in\left\{0;\dfrac{-1}{2}\right\}\)

a) Số hạng thứ 20 (n=20) là 

\(\left(20-1\right).4=76\)

\(A=1-5+9-13+17-21+...+76\)

\(A=\left(-4\right)+\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)

\(A=\left(-4\right).38=-152\)

b) Số hạng thứ n là:

\(\left(n-1\right).4\)

\(\)\(A=1-5+9-13+17-21+...+\left(n-1\right).4\)

\(A=\left(-4\right)+\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)   ((n-1).2 số -4)

\(A=\left(-4\right).\left(n-1\right).2=-8\left(n-1\right)\)

điên à

\(\dfrac{5}{7}:x+1=\dfrac{-15}{30}\)

\(\dfrac{5}{7}:x=\dfrac{-15}{30}-1\)

\(\dfrac{5}{7}:x=\dfrac{-45}{30}\)

\(x=\dfrac{5}{7}:\dfrac{-45}{30}\)

\(x=\dfrac{-10}{21}\)

\(\dfrac{5}{7}:x+1=-\dfrac{15}{30}\)

\(\Rightarrow\dfrac{5}{7}:x+1=-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{5}{7}:x=-\dfrac{1}{2}-1\)

\(\Rightarrow\dfrac{5}{7}:x=-\dfrac{3}{2}\)

\(\Rightarrow x=\dfrac{5}{7}:-\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{10}{21}\)

\(\left(x-\dfrac{1}{2}\right):\dfrac{9}{11}=\dfrac{11}{3}\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{11}{3}\cdot\dfrac{9}{11}\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{9}{3}\)

\(\Rightarrow x-\dfrac{1}{2}=3\)

\(\Rightarrow x=3+\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{7}{2}\)

\(\left(x-\dfrac{1}{2}\right):\dfrac{9}{11}=\dfrac{11}{3}\\ \\ \\ \Rightarrow x-\dfrac{1}{2}=\dfrac{11}{3}\cdot\dfrac{9}{11}=\dfrac{9}{3}=3\\ \\ \\ \Rightarrow x=3+\dfrac{1}{2}=3\dfrac{1}{2}\)

\(\left(\dfrac{9}{2}-2x\right)\cdot1\dfrac{4}{7}=\dfrac{11}{14}\\ \\\\ \Rightarrow\left(\dfrac{9}{2}-2x\right)\cdot\dfrac{11}{7}=\dfrac{11}{14}\\ \\\\ \Rightarrow\dfrac{9}{2}-2x=\dfrac{11}{14}:\dfrac{11}{7}=\dfrac{11}{14}\cdot\dfrac{7}{11}=\dfrac{1}{2}\\ \\ \\ \Rightarrow2x=\dfrac{9}{2}-\dfrac{1}{2}=\dfrac{8}{2}=4\\ \\ \\ \Rightarrow x=4:2=2\)

(\(\dfrac{9}{2}\) - 2\(x\)) \(\times\) 1\(\dfrac{4}{7}\) = \(\dfrac{11}{14}\)

( \(\dfrac{9}{2}\) - 2\(x\))  \(\times\) \(\dfrac{11}{7}\) =   \(\dfrac{11}{14}\)

\(\dfrac{9}{2}\)  - 2\(x\)              = \(\dfrac{11}{14}\) : \(\dfrac{11}{7}\)

\(\dfrac{9}{2}\) - 2\(x\)               = \(\dfrac{1}{2}\)

       2\(x\)               =   \(\dfrac{9}{2}\) - \(\dfrac{1}{2}\)

       2\(x\)               = 4

         \(x\)                = 4 : 2

          \(x\)               = 2

Theo đề ta có y = ax, x = bz. Suy ra y = ax = abz.

Vậy y tỉ lệ thuận với z theo hệ số tỉ lệ ab

\(-\dfrac{17}{14}:\left(-\dfrac{34}{7}\right)+\dfrac{10}{3}\left(\dfrac{1}{5}-\dfrac{3}{4}\right)-\dfrac{6}{5}\left(\dfrac{1}{3}-\dfrac{5}{6}\right)\)

\(=\dfrac{17}{14}.\dfrac{7}{34}+\dfrac{10}{3}.\left(-\dfrac{11}{20}\right)-\dfrac{6}{5}.\left(-\dfrac{9}{18}\right)\)

\(=\dfrac{1}{4}-\dfrac{11}{6}+\dfrac{9}{15}\)

\(=\dfrac{15-110+36}{60}\)

\(=-\dfrac{59}{60}\)

Ta thấy chỉ có \(n=3\) thỏa mãn đẳng thức \(\left(n+5\right)^2=64\left(n-2\right)^3\)vì

- \(\left(n+5\right)^2\) là 1 số chính phương

- \(64\) là 1 số chính phương

- \(\left(n-2\right)^3\) không phải số chính phương

- \(\)\(\left(n+5\right)^2< 64\left(n-2\right)^3,\forall n>3\)